When an object goes up we say that it gained potential energy but it is doing positive work on the earth so it should lose energy.Please correct me.

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    $\begingroup$ Who's doing the work? $\endgroup$ – user191954 Oct 17 '18 at 16:02
  • $\begingroup$ The object which is going up is doing the work. $\endgroup$ – user64348 Oct 18 '18 at 8:27
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    $\begingroup$ @user64348 Don't you mean something or someone is doing work against gravity to raise the object? $\endgroup$ – Bob D Oct 21 '18 at 22:40

When an object goes up we say that it gained potential energy


but it is doing positive work on the earth

Make sure you get the signs right. Potential energy is negative mgh.

The positive work you do is offset by the negative PE. The total energy is zero before, during and after the move.


When an object goes up in the earth’s gravitational field it loses KE and gains PE. The work done on the object is negative because the force is in the opposite direction of the motion.

By Newton’s third law there is an equal and opposite force on the earth. However, because the earth does not move no work is done on the earth.


When an object of mass $m$ goes up near the surface of the earth, something or somebody else is doing work on the object to raise it. The object itself is not doing work on the earth and it is not losing energy. The something or somebody doing the work on the object against gravity to raise it is losing energy, but it is transferring that energy to the object in the form of increased gravitational potential energy.

There is no need for there to be a net increase or decrease in kinetic energy. The key is the object needs to begin at rest and end at rest. In order to accomplish this, we initially need to apply an external upward force, $F_{ext}$, slightly greater than $mg$, to give it small upward acceleration $a$. Let’s say we do this for a brief time $dt$ and therefore over a short distance $dh$. The mass thereby attains a small velocity $v=adt$, a small increase in kinetic energy of $ ½ m(adt)^2 $ and a small increase in the potential energy of the mass m of $(mg) dh$. We now immediately reduce our upward force so that it equals the downward gravitational force. The mass is now rising at constant velocity v and so there is no subsequent change in KE, however its potential energy keeps increasing since the mass continues to rise. The work to accomplish this increase in potential energy is due to our constant application of an upward force equal to the force of gravity.

As we approach point 2 we are still left with the small kinetic energy. Therefore, prior to reaching point 2, we reduce the external upward force, $F_{ext}$, to slightly less than $mg$, to give it a small negative acceleration. We do this for sufficient time to bring the mass to rest at point 2 a height $h$ above point 1. During this period gravity now does a small amount of negative work resulting in a change in kinetic energy of $ -½ m(adt)^2 $. Consequently the total change in kinetic energy going from point 1 to point 2 is zero. Since there is no loss in height during this period, there is no loss in potential energy.

The end result in going from 1 to 2 is an increase in gravitational potential energy of

$$ \int_1^2 (mg)dh = (mg)h$$.

with no net change in kinetic energy. This increase in potential energy comes, of course, came from the external agent that did work on the object.

Hope this helps.


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